Consider the following short exact sequence of groups
$$ 0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{C} \longrightarrow \mathbb{C}^{\ast} \longrightarrow 1 $$ where the last map is given by $z \mapsto \exp(2 i \pi z)$. How can one show that this sequence is not split?
It is easy to show that a continuous section cannot exist but here a section should just be a morphism.
Any comment or answer would be highly appreciated! Thanks!
Let $f:\mathbb{C}^*\to\mathbb{C}$ be a group homomorphism and $K=\{z\in\mathbb{C}^*\ |\ z^n=1\text{ for some }n>0\}$ be the subgroup of all elements of finite order, which is an infinite subgroup of $\mathbb{C}^*$. Since $\mathbb{C}$ has no non-trivial elements of finite order, then $f(K)=0$. And so no homomorphism $g:\mathbb{C}\to\mathbb{C}^*$ can give us $g\circ f=\mathrm{id}_{\mathbb{C}^*}$.
This can be generalized as follows: given a short exact sequence
$$0\to A\to B\to C\to 0$$
it cannot be split, if $B$ is torsion-free while $C$ is not.
Or even more: it cannot be split if $C$ contains a nontrivial element $m\in C$ such that every morphism $C\to B$ maps it to zero.